Let $a_0$, $a_1$, $a_2$, $\dots$ be an infinite sequence of real numbers such that $a_0 = \frac{5}{13}$ and
\[
  a_{n} = 2 a_{n-1}^2 - 1
\]for every positive integer $n$.  Let $c$ be the smallest number such that for every positive integer $n$, the product of the first $n$ terms satisfies the inequality
\[|a_0 a_1 \dotsm a_{n - 1}| \le \frac{c}{2^n}.\]What is the value of $100c$, rounded to the nearest integer?
Define the sequence $(\theta_n)$ by $\theta_0 = \arccos \frac{5}{13}$ and
\[\theta_n = 2 \theta_{n - 1}.\]Then $\cos \theta_0 = \frac{5}{13},$ and
\begin{align*}
\cos \theta_n &= \cos (2 \theta_{n - 1}) \\
&= 2 \cos^2 \theta_{n - 1} - 1.
\end{align*}Since the sequences $(a_n)$ and $(\cos \theta_n)$ have the same initial term, and the same recursion, they coincide.

We have that
\[\sin^2 \theta_0 = 1 - \cos^2 \theta_0 = \frac{144}{169}.\]Since $\theta_0$ is acute, $\sin \theta_0 = \frac{12}{13}.$

Now,
\begin{align*}
a_0 a_1 \dotsm a_{n - 1} &= \cos \theta_0 \cos \theta_1 \dotsm \cos \theta_{n - 1} \\
&= \cos \theta_0 \cos 2 \theta_0 \dotsm \cos 2^{n - 1} \theta_0.
\end{align*}Multiplying both sides by $\sin \theta_0 = \frac{12}{13},$ we get
\begin{align*}
\frac{12}{13} a_0 a_1 \dotsm a_{n - 1} &= \sin \theta_0 \cos \theta_0 \cos 2 \theta_0 \cos 4 \theta_0 \dotsm \cos 2^{n - 1} \theta_0 \\
&= \frac{1}{2} \sin 2 \theta_0 \cos 2 \theta_0 \cos 4 \theta_0 \dotsm \cos 2^{n - 1} \theta_0 \\
&= \frac{1}{4} \sin 4 \theta_0 \dotsm \cos 2^{n - 1} \theta_0 \\
&= \dotsb \\
&= \frac{1}{2^n} \sin 2^n \theta_0.
\end{align*}Hence,
\[|a_0 a_2 \dotsm a_{n - 1}| = \frac{1}{2^n} \cdot \frac{13}{12} |\sin 2^n \theta_0| \le \frac{1}{2^n} \cdot \frac{13}{12}.\]This tells us $c \le \frac{13}{12}.$

We can compute that $a_1 = 2a_0^2 - 1 = 2 \left( \frac{5}{13} \right)^2 - 1 = -\frac{119}{169},$ so
\[\frac{5}{13} \cdot \frac{119}{169} \le \frac{c}{4}.\]Then $c \ge \frac{2380}{2197}.$  The bound
\[\frac{2380}{2197} \le c \le \frac{13}{12}\]tells us that the integer closest to $100c$ is $\boxed{108}.$